Subject: Re: Russell's Paradox Resolved
From: Cattabriga 
Date: Thu, 28 Sep 2000 19:47:19 GMT
Newsgroups: sci.logic,sci.math
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References: <[email protected]>
X-Article-Creation-Date: Thu Sep 28 19:47:19 2000 GMT



In article <[email protected]>,
  [email protected] wrote:

>
> Therefore  6. ~({x:~(x e x)} e {x:~(x e x)},  is a theorem.
>


Your all-first-order way to try a solution seems to me
not too far from the statement that \neg EyAx(x \in y <-> x\notin x),
is a first order theorem. Of course the statement is true.

I think instead we should follow Poincare' "... not predicative
definitions can not be substituted to _ defined terms _. In this
condition, logistics is no longer sterile: it generates contradictions."
(Science et methode, 1902) The inquiry for the solution of Russell's
antinomy, which affects comprehension axiom schema, should follow the
conditions established by the theory of definition (See Padoa, Beth,
Craig, Mckinsey, Tarski, ...Suppes ...).
Comprehension axiom schema was originally conceived as a schema for
defining sets, so we are legitimate to consider it under the point of
view of the theory of definition.
Under this perspective in defining a new operation symbol (or a new
individual constant, i.e. an operation symbol of rank zero)
we are required to have a preceding theorem which guarantees that the
operation is uniquely defined. If the restriction on the uniqueness is
dropped then a contradiction can be derived. (See for example page 159,
Introduction to Logic, Patrick Suppes, Dover Publications, 1999).

For comprehension axiom schema the uniqueness is ensured by the axiom of
extensionality.

COMPREHENSION =      Az_1...Az_nEyAx(x \in y <-> P(x)),
where z_1,...,z_n are the free variables of P(x) other than x,
and y is not a free variable of P(x).

EXTENSIONALITY =      AxAy[Az(z \in x <-> z \in y) -> x = y]


Extensionality, without any additional axioms, obviously implies that
for every condition P(x) on x (in COMPREHENSION) there exists "at most"
one set y which contains exactly those elements x which fulfil the
condition P(x); in other words, if y_1 and y_2 are two sets each of
which contains exactly those elements x which fulfil the condition P(x),
then y_1 and y_2 are equal.

Usually Russell's antinomy is as follows.

Russell's antinomy argumentation. There exists no set (element) which
contains exactly those elements which do not contain
themselves (in symbols \neg EyAx(x \in y <-> x\notin x)).

Proof. By contradiction. Assume that y is a set such that for every
element x, x \in y if and only if x \notin x. For x=y, we have y \in y
if and only if y \notin y. Since, obviously, y \in y or y \notin y, and
as we saw, each of y \in y and y \notin y implies the other statement, we
have both y \in y and y \notin y, which is a contradiction.


This proof holds of course as a first order theorem, but it does not
hold for a whatever system based on COMPREHENSION and EXTENSIONALITY.
This is indeed the simple solution of Russell's antinomy!
In other words: Russell's antinomy has never affected Frege system!

It is very easy to show that by extensionality the
above proof cannot be accomplished. If by comprehension we define
x \in y if and only if x \notin x then by extensionality
(x \in y if and only if x \notin x) -> \neg(x = y) and hence it is never
the case that x = y. This is the uniqueness condition. Thence the above
proof cannot be accomplished since "x=y" cannot be assumed and y \in y
if and only if y \notin y cannot be derived.

We can then state that Russell's antinomy argumentation does not hold
in a system based on COMPREHENSION and EXTENSIONALITY, because it
violate the restriction on the uniqueness of y established by
extensionality.

In other terms. For a simple first order rule from Ax(x \in a <-> x
\notin x) we can get (a \in a <-> a \notin a). But this rule is not
valid in  a set theory where, by extensionality,
(x \in a <-> x \notin x) -> \neg(x = a).



Opinions?


Paola Cattabriga.

______________________________________________________________

http://digilander.iol.it/PC61
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